Giải thích các bước giải:
$I=\int\dfrac{1}{1+2\cos^2x}dx$
$\to I=\int\dfrac{1}{\cos(2x)+2}dx$
$\to I=\int\dfrac{1}{2(\cos(2x)+2)}d(2x)$
Đặt $v=\tan x$
$\to I=\dfrac 12.\int\dfrac{2}{v^2+3}dv$
$\to I=\dfrac{1}{2}\cdot \:2\cdot \dfrac{1}{\sqrt{3}}\arctan \left(\dfrac{v}{\sqrt 3}\right)$
$\to I=\dfrac{1}{\sqrt{3}}\arctan \left(\dfrac{\tan \left(x\right)}{\sqrt{3}}\right)+C$
$\to \int^{\dfrac{\pi}{4}}_{-\dfrac{\pi}{4}}\dfrac{1}{1+2\cos^2x}dx=\dfrac{\pi }{3\sqrt{3}}$
